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I have tried to solve this limit with L'Hopital's rule; $\lim_{x\to0+}(1+x)^{lnx}$ but i applied several times with any result. I would appreciate if somebody can help me. Note: Wolfram alpha said that the result is 1, but it doesn't appear the steps to get it.

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  • $\begingroup$ I have merged Dr Sonnhard Graubner's answers. If you want to accept his answer, you will need to accept it again. $\endgroup$ – robjohn Oct 6 '14 at 17:47
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hint: write $(1+x)^{\ln(x)}$ as $e^{\frac{\ln(1+x)}{\frac{1}{\ln(x)}}}$

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  • $\begingroup$ I did it, but i applied L'Hopital's rule several times, and the indeterminate doesn't disappear $\endgroup$ – egarro Oct 6 '14 at 17:04
  • $\begingroup$ the first step is $\frac{\frac{1}{1+x}}{\frac{-1}{x\ln(x)^2}}$ and then $-\frac{x\ln(x)^2}{1+x}$ $\endgroup$ – Dr. Sonnhard Graubner Oct 6 '14 at 17:14
  • $\begingroup$ And then we get the form 0*infinity, right? $\endgroup$ – egarro Oct 6 '14 at 17:22
  • $\begingroup$ yes and then we write $-\frac{\ln(x)^2}{1+\frac{1}{x}}$ and the next step is the last one $\frac{2\ln(x)\frac{1}{x}}{-\frac{1}{x^2}}$ $\endgroup$ – Dr. Sonnhard Graubner Oct 6 '14 at 17:28
  • $\begingroup$ Thank so much! I already get the result. $\endgroup$ – egarro Oct 6 '14 at 17:35
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To apply l'Hopital's rule, you first need to get the something in the form $\lim_{x\to a}\frac{f(x)}{g(x)}$, where $f(x)$ and $g(x)\to 0$ as $x\to a$. In you case, if we let $L=\lim_{x\to0^+}(1+x)\ln x$, then taking $\ln$ of both sides, we get $$ \ln L = \lim_{x\to 0^+}\ln x\ln (1+x)=\lim_{x\to 0^+}\frac{\ln (1+x)}{1/\ln x} $$ Now, you can apply l'Hopital's rule to the last expression, allowing you to find $\ln L$, then raise $e$ to that to find $L$.

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  • $\begingroup$ I did that, but i applied the rule like three times, with any result. $\endgroup$ – egarro Oct 6 '14 at 17:00
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Without using Hospital rule also we can evaluate this limit.
Substitute $x=e^t$ then $x\to0^+$ is same as $t\to -\infty.$ $$\lim_{x\to0^+}(1+x)^{\ln x}=\lim_{t\to-\infty} (1+e^t)^t=\lim_{t\to-\infty} (1+O(e^t))=1.$$

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  • $\begingroup$ How did the $t$ (which goes to $-\infty$) disappear from the exponent? $\endgroup$ – robjohn Oct 6 '14 at 17:17
  • $\begingroup$ @robjohn: We can use the binomial theorem. $\endgroup$ – Bumblebee Oct 6 '14 at 17:21
  • $\begingroup$ With an exponent approaching $-\infty$? That seems hard to do. $\endgroup$ – robjohn Oct 6 '14 at 17:51
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Hint: $f(x,y)=x^y$ is continuous at $(e,0)$ so $$ \begin{align} \lim_{x\to0}\left[\left(1+x\right)^{1/x}\right]^{x\log(x)} &=\left[\lim_{x\to0}\left(1+x\right)^{1/x}\right]^{\lim\limits_{x\to0}x\log(x)}\\ \end{align} $$ You can also write $x\log(x)=\dfrac{\log(x)}{1/x}$ and use L'Hospital once.


Alternate Approach $$ \begin{align} \lim_{x\to0}\log\left(\left(1+x\right)^{\log(x)}\right) =\lim_{x\to0}\frac{\log(x)}{1/x}\lim_{x\to0}\frac{\log(1+x)}{x} \end{align} $$ Each limit requires one application of L'Hospital.

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